(k+3)(k+9)=0

Solution for (k+3)(k+9)=0 equation:

(k+3)(k+9)=0

We multiply parentheses ..

(+k^2+9k+3k+27)=0

We get rid of parentheses

k^2+9k+3k+27=0

We add all the numbers together, and all the variables

k^2+12k+27=0

a = 1; b = 12; c = +27;
Δ = b2-4ac
Δ = 122-4·1·27
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$k_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6}{2*1}=\frac{-18}{2}
=-9
$
$k_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6}{2*1}=\frac{-6}{2}
=-3
$